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()
22
ˆˆ
3.0 [(3.0)( 6.0) (8.0)(5.0)]k ( 1.7 10 kg m s)k.
=−
−=
−
×
⋅
G
A
(b) The torque is given by Eq. 1114,
G
G
G
τ
=×
rF
.
We write
G
rxy
=+
##
i
j and
G
FF
x
=
#
i and
obtain
G
=+×
=
−
xy
F
y
F
xx
#
#
ij
i
k
e
j
e
j
since
ii
×=
0
and
## #
ji k
.
×=−
Thus, we find
(
)
8.0m
7.0N k (56N m)k.
−
=
⋅
G
(c) According to Newton’s second law
G
G
A
=
dd
t
, so the rate of change of the angular
momentum is 56 kg
⋅
m
2
/s
2
, in the positive
z
direction.
33. If we write (for the general case)
G
rxyz
=++
###
k
,
then (using Eq. 330) we find
G
G
r
v
×
is equal to
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics, Angular Momentum, Momentum

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